Lecture7 & 8_FDM

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Multiplexing Data Communication and Networks Prof. Dr. Abdul Qadeer Khan Rajput By

Transcript of Lecture7 & 8_FDM

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Multiplexing

Data Communication and Networks

Prof. Dr. Abdul Qadeer Khan Rajput

By

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Multiplexing is a set of techniques that allows the simultaneous transmission of multiple signals across a single data link.

Multiplexer combines the incoming lines into one [single stream].

De-multiplexer (DEMUX) separates the stream back into its component transmissions (one to many) and directs them to their corresponding lines.

Multiplexing

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Basic Definitions

Multiplexer(MUX): device to combine signals to one composite signal.

Demultiplexer(DEMUX): device to separate the composite signal into signal components

Path: refers to the physical link. Channel: refers to a portion of a path.

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Types of Multiplexing

There are two basic types of multiplexing in use: Frequency Division Multiplexing (FDM) & Time Division Multiplexing (TDM).

Wavelength Division Multiplexing (WDM) is one special kinds of FDM used in fiber optical transmission.

TDM can be further divided into synchronous TDM and asynchronous TDM.

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Categories of multiplexing

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FDM Definition Frequency Division Multiplexing (FDM)

is the technique to combine signals from different bands.

Each signal modulates a separate carrier frequency.

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FDM Multiplexing – time domain

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FDM Demultiplexing – time domain

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FDM Multiplexing – frequency domain

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FDM Demultiplexing – frequency domain

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Concept of Guard Band

To prevent channel overlapping, they must be separated by unused bandwidth called Guard Band.

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Figure Dividing a link into channels

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Frequency Division Multiplexing

Frequency-division multiplexing is an analog technique that can be applied when the bandwidth of a link (in hertz) is greater than the combined bandwidths of the signals to be transmitted.

Signals generated by each sending device modulate different carrier frequencies.

FDM is an analog multiplexing technique that combines signals

Figure FDM

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Frequency Division Multiplexing

FDM Useful bandwidth of

medium exceeds required bandwidth of channel

Each signal is modulated to a different carrier frequency

Carrier frequencies separated so signals do not overlap (guard bands)

e.g. broadcast radio Channel allocated even if

no data

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17Figure FDM process

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FDM System

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FDM Example

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Analog Carrier Systems

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Example 1

Assume that a voice channel occupies a bandwidth of 4 KHz. We need to combine three voice channels into a link with a bandwidth of 12 KHz, from 20 to 32 KHz. Show the configuration using the frequency domain without the use of guard bands.

SolutionSolution

Shift (modulate) each of the three voice channels to a different bandwidth, as shown in Figure .

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22Figure Example 1

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Five channels, each with a 100-KHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 KHz between the channels to prevent interference?

SolutionSolutionFor five channels, we need at least four guard bands. This means that the required bandwidth is at least 5 x 100 + 4 x 10 = 540 KHz, as shown in Figure.

Example 2

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Figure Example 2

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Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration using FDM

SolutionSolutionThe satellite channel is analog. We divide it into four channels, each channel having a 250-KHz bandwidth. Each digital channel of 1 Mbps is modulated such that each 4 bits are modulated to 1 Hz. One solution is 16-QAM modulation. Figure 6.8 shows one possible configuration.

Example 3

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26Figure Example 3

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In analog hierarchy, 12 voice channels are multiplexed onto a higher-bandwidth line to create a group.

The join them as group, supergroup, master group and jumbo group.

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Analog hierarchy

Figure Analog hierarchy

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The Advanced Mobile Phone System (AMPS) uses two bands. The first band, 824 to 849 MHz, is used for sending; and 869 to 894 MHz is used for receiving. Each user has a bandwidth of 30 KHz in each direction. The 3-KHz voice is modulated using FM, creating 30 KHz of modulated signal. How many people can use their cellular phones simultaneously?

SolutionSolution

Each band is 25 MHz. If we divide 25 MHz into 30 KHz, we get 833.33. In reality, the band is divided into 832 channels.

Example 4

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Four 1-Kbps connections are multiplexed together. A unit is 1 bit. Find (1) the duration of 1 bit before multiplexing, (2) the transmission rate of the link, (3) the duration of a time slot, and (4) the duration of a frame?

SolutionSolution

We can answer the questions as follows:1. The duration of 1 bit is 1/1 Kbps, or 0.001 s (1 ms).2. The rate of the link is 4 Kbps.3. The duration of each time slot 1/4 ms or 250 s. 4. The duration of a frame 1 ms.

Example 5

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TDM Definition

Time Division Multiplex (TDM) is the process of combining signals together in the time domain.

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TDM Basics The basic principle of

TDM is Sampling Theorem.

The basic unit in TDM is Time Slot.

The data in TDM is grouped into Frames, which consists of one cycle of time slots.

T

3T

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TDM

TDM is a digital multiplexing technique to combine data

Instead of sharing a portion of the bandwidth as in FDM, time is shared.

Each connection occupies a portion of time in the link.

In a TDM, the data rate of the link is n times faster, and the unit duration is n times shorter

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Interleaving

Interleaving is the process of multiplexing. In TDM, synchronization between the

sender and receiver is very important.

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Interleaving

TDM can be visualized as two fast rotating switches, one on the multiplexing side and the other on the de-multiplexing side. The switches are synchronized and rotate at the same speed, but in opposite directions. On the multiplexing side, as the switch opens in front of a connection, that connection has the opportunity to send a unit onto the path. This process is called interleaving.

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Four channels are multiplexed using TDM. If each channel sends 100 bytes/s and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the duration of a frame, the frame rate, and the bit rate for the link.

SolutionSolutionThe multiplexer is shown in Figure.

Example 6

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T-1 line for multiplexing telephone lines

T lines are digital lines designed for the transmission of digital data, audio, or video.

T lines also can be used for analog transmission (regular telephone connections), provided the analog signals are sampled first, then time-division multiplexed.

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A multiplexer combines four 100-Kbps channels using a time slot of 2 bits. Show the output with four arbitrary inputs. What is the frame rate? What is the frame duration? What is the bit rate? What is the bit duration?

SolutionSolutionFigure shows the output for four arbitrary inputs.

Example 7

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Framing bits

For synchronization between multiplexer and demultiplexer, one or more synchronization bits are usually added to the beginning of each frame. These bits, called framing bits, follow a pattern, frame to frame, that allows the demultiplexer to synchronize with the incoming stream so that it can separate the time slots accurately.In bit padding, the multiplexer adds extra bits to a device’s source stream to force the speed relationships among the various devices into integer multiples of each other.

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We have four sources, each creating 250 characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (1) the data rate of each source, (2) the duration of each character in each source, (3) the frame rate, (4) the duration of each frame, (5) the number of bits in each frame, and (6) the data rate of the link.

Example 8

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We can answer the questions as follows:

1. The data rate of each source is 2000 bps = 2 Kbps.2. The duration of a character is 1/250 s, or 4 ms.3. The link needs to send 250 frames per second.4. The duration of each frame is 1/250 s, or 4 ms. 5. Each frame is 4 x 8 + 1 = 33 bits.6. The data rate of the link is 250 x 33, or 8250 bps.

SolutionSolution

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Two channels, one with a bit rate of 100 Kbps and another with a bit rate of 200 Kbps, are to be multiplexed. How this can be achieved? What is the frame rate? What is the frame duration? What is the bit rate of the link?

SolutionSolutionWe can allocate one slot to the first channel and two slots to the second channel. Each frame carries 3 bits. The frame rate is 100,000 frames per second because it carries 1 bit from the first channel. The frame duration is 1/100,000 s, or 10 ms. The bit rate is 100,000 frames/s x 3 bits/frame, or 300 Kbps.

Example 9

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Digital Signal (DS) hierarchy

DS-0 service is a single digital channel of 64 Kbps.

DS-1 is a 1.544-Mbps service [24 times 64 Kbps plus 8Kbps of overhead]

And so on. DS-0, DS-1, and so on

are the names of services. To implement those services, the telephone companies use T lines (T-1 to T-4).

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Table DS and T lines ratesTable DS and T lines rates

Service LineRate

(Mbps)Voice

Channels

DS-1DS-1 T-1T-1 1.5441.544 2424

DS-2DS-2 T-2T-2 6.3126.312 9696

DS-3DS-3 T-3T-3 44.73644.736 672672

DS-4DS-4 T-4T-4 274.176274.176 40324032

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Table . Table . E line E line ratesrates

E LineRate

(Mbps)Voice

Channels

E-1E-1 2.0482.048 3030

E-2E-2 8.4488.448 120120

E-3E-3 34.36834.368 480480

E-4E-4 139.264139.264 19201920

Europeans use a version of T lines called E lines The two systems are conceptually identical, but their

capacities differ.

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Synchronous TDM

In synchronous TDM, each source is pre-assigned a fixed location of time slot.

Each source can and only can send information at the time slot given to it. If a source has no data to send, its time slot remains empty. This can lead to inefficiency.

If n sources are grouped together, the total data rate of the path is n times the original data rate of each source.

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Synchronous TDM Example

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Synchronous TDM - Multiplexing

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Synchronous TDM - Demultiplexing

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We have four sources, each creating 250 characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (1) the data rate of each source, (2) the duration of each character in each source, (3) the frame rate, (4) the duration of each frame, (5) the number of bits in each frame, and (6) the data rate of the link.

TDM Example

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We can answer the questions as follows:

1. The data rate of each source is (250 chars x 8 bits) = 2000 bps = 2 kbps.

2. The duration of a character is 1/250 s, or 4 ms.3. The link needs to send 250 frames per second.4. The duration of each frame is 1/250 s, or 4 ms. 5. Each frame is (4sources x 8bits) + 1sync = 33 bits.6. The data rate of the link is 250 x 33, or 8250 bps.

(strictly, ‘bit rate’ of the link is 8250bps – the actual data rate is only 8000 data bits/s)

SolutionSolution

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Asynchronous TDM In asynchronous TDM or statistical

TDM, only sources containing data will be sent with time slot. Therefore, asynchronous TDM can avoid bandwidth waste in synchronous TDM.

But, in order to distinguish data from different sources, address should be added into the frame structure, increasing the overhead of the transmission.

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Asynchronous TDM Example

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Address and Overhead As shown below, address is added before the

data from each source. It is practical only when the data size for each

time slot is relatively larger than the address.

In this example, the addressing information takes up as much space as the actual data - inefficient

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TDM in Telephone System FDM was used in the original telephone

system, but now the telephone lines (except the subscriber line) are all in digital form, so FDM is not in use now.

There are two types of TDM used in telephone system. In US, T-1 line with basic rate of 1.544 Mbps is used, while in Europe and China, E-1 Line of 2.048 Mbps is in use.

DS is a ‘service’, T-n is a ‘line’

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Digital Signal (DS) Hierarchy in US

8k sample/sec x 8bit samples gives 64kbps basic line

DS1 = 24 x DS0 = 24x64k=1536k

1536k + 8k(sync) =1544k

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T-1 Frame Structure

Data sync

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One frame = 24 * 8 + 1 sync bit

Frame rate = 193 * 8000 = 1.544 Mbps

Sync bit